Codeforces Round #346 (Div. 2) 解题报告

659A. Round House

Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to n. Entrance n and entrance 1 are adjacent.

Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance a and he decided that during his walk he will move around the house b entrances in the direction of increasing numbers (in this order entrance n should be followed by entrance 1). The negative value of b corresponds to moving |b| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance n). If b = 0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.

Input
The single line of the input contains three space-separated integers n, a and b (1 ≤ n ≤ 100, 1 ≤ a ≤ n,  - 100 ≤ b ≤ 100) — the number of entrances at Vasya’s place, the number of his entrance and the length of his walk, respectively.

Output
Print a single integer k (1 ≤ k ≤ n) — the number of the entrance where Vasya will be at the end of his walk.

input
6 2 -5
output
3
input
5 1 3
output
4
input
3 2 7
output
3

题目链接：cf-659A

题目大意：n个数从小到大排列成环，一个人起始位置在a，走b步，如b为正，则向数字增大地方走，反之向分数小的地方走。

以下是代码：

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
    int n,a,b;
    cin >> n >> a >> b;
    if(b > 0)
    {
        int ans = (a + b) % n;
        if (ans == 0) ans = n;
        cout << ans << endl;
    }
    else
    {
        b = abs(b) % n;
        if (b == 0) b = n;
        if (a > b)
        {

            cout << (a - b) << endl;
        }
        else
        {
            cout << n - (b - a)<< endl;
        }
    }
    return 0;
}

659B. Qualifying Contest

Very soon Berland will hold a School Team Programming Olympiad. From each of the m Berland regions a team of two people is invited to participate in the olympiad. The qualifying contest to form teams was held and it was attended by n Berland students. There were at least two schoolboys participating from each of the m regions of Berland. The result of each of the participants of the qualifying competition is an integer score from 0 to 800 inclusive.

The team of each region is formed from two such members of the qualifying competition of the region, that none of them can be replaced by a schoolboy of the same region, not included in the team and who received a greater number of points. There may be a situation where a team of some region can not be formed uniquely, that is, there is more than one school team that meets the properties described above. In this case, the region needs to undertake an additional contest. The two teams in the region are considered to be different if there is at least one schoolboy who is included in one team and is not included in the other team. It is guaranteed that for each region at least two its representatives participated in the qualifying contest.

Your task is, given the results of the qualifying competition, to identify the team from each region, or to announce that in this region its formation requires additional contests.

Input
The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 10 000, n ≥ 2m) — the number of participants of the qualifying contest and the number of regions in Berland.

Next n lines contain the description of the participants of the qualifying contest in the following format: Surname (a string of length from 1 to 10 characters and consisting of large and small English letters), region number (integer from 1 to m) and the number of points scored by the participant (integer from 0 to 800, inclusive).

It is guaranteed that all surnames of all the participants are distinct and at least two people participated from each of the m regions. The surnames that only differ in letter cases, should be considered distinct.

Output
Print m lines. On the i-th line print the team of the i-th region — the surnames of the two team members in an arbitrary order, or a single character “?” (without the quotes) if you need to spend further qualifying contests in the region.

input
5 2
Ivanov 1 763
Andreev 2 800
Petrov 1 595
Sidorov 1 790
Semenov 2 503
output
Sidorov Ivanov
Andreev Semenov
input
5 2
Ivanov 1 800
Andreev 2 763
Petrov 1 800
Sidorov 1 800
Semenov 2 503
output
?
Andreev Semenov

题目链接：cf-659B

题目大意：n个人，m个区。给出n个人的姓名（保证不相同），属于的区域，所得分数。从每个区域中选出成绩最好的两个人去参加比赛，输出这两个人的名字。如果第三个人的成绩和第二个人的成绩相同，则输出“？”，保证结果不确定。

题目思路：利用vector，将每个区的人分别存储好。然后进行比较

以下是代码：

#include <bits/stdc++.h>
using namespace std;
struct node
{
    int score;
    string name;
};
vector <node> vec[100005];
bool cmp(node a,node b)
{
    return a.score > b.score;
}
int main()
{
    int n,m;
    cin >> n >> m;
    for (int i = 0; i < n; i++)
    {
        node tmp;
        int no;
        cin >> tmp.name >> no >> tmp.score;
        vec[no].push_back(tmp);
    }

    for (int i = 1; i <= m; i++)
    {
        sort(vec[i].begin(),vec[i].end() , cmp);
        if(vec[i].size() == 1)
        {
            cout << "?" << endl;
        }
        else if (vec[i].size() == 2)
        {
            node one = vec[i][0];
            node two = vec[i][1];
            cout << one.name << " " << two.name << endl;
        }
        else
        {
            node one = vec[i][0];
            node two = vec[i][1];
            node three = vec[i][2];
            if (three.score == two.score) cout << "?" << endl;
            else cout << one.name << " " << two.name << endl;
        }
    }
    return 0;
}

659C. Tanya and Toys

In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the i-th type costs i bourles.

Tania has managed to collect n different types of toys a1, a2, …, an from the new collection. Today is Tanya’s birthday, and her mother decided to spend no more than m bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.

Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.

Input
The first line contains two integers n (1 ≤ n ≤ 100 000) and m (1 ≤ m ≤ 109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.

The next line contains n distinct integers a1, a2, …, an (1 ≤ ai ≤ 109) — the types of toys that Tanya already has.

Output
In the first line print a single integer k — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed m.

In the second line print k distinct space-separated integers t1, t2, …, tk (1 ≤ ti ≤ 109) — the types of toys that Tanya should choose.

If there are multiple answers, you may print any of them. Values of ti can be printed in any order.

input
3 7
1 3 4
output
2
2 5
input
4 14
4 6 12 8
output
4
7 2 3 1

题目链接：cf-659C

题目大意：总共有10^9种礼品，价格为1~10^9，女孩已经有了n种，妈妈答应用m元去买礼品给女孩，女孩想要尽可能多的礼品，已经有的不要，问最多买几个

题目思路：想了很久优化，后来发现直接暴力即可

以下是代码：

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
int a[100005];
map <int,int> mp;
vector <int> ans;
int main()
{
    int n,m;
    cin >> n >> m;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        mp[a[i]] = 1;
    }
    for (int i = 1;; i++)
    {
        if (m < i) break;
        if (mp[i] == 0)
        {
            ans.push_back(i);
            m -= i;
        }
    }
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); i++)
    {
        cout << ans[i] << " ";
    }
    return 0;
}

659D. Bicycle Race

Maria participates in a bicycle race.

The speedway takes place on the shores of Lake Lucerne, just repeating its contour. As you know, the lake shore consists only of straight sections, directed to the north, south, east or west.

Let’s introduce a system of coordinates, directing the Ox axis from west to east, and the Oy axis from south to north. As a starting position of the race the southernmost point of the track is selected (and if there are several such points, the most western among them). The participants start the race, moving to the north. At all straight sections of the track, the participants travel in one of the four directions (north, south, east or west) and change the direction of movement only in bends between the straight sections. The participants, of course, never turn back, that is, they do not change the direction of movement from north to south or from east to west (or vice versa).

Maria is still young, so she does not feel confident at some turns. Namely, Maria feels insecure if at a failed or untimely turn, she gets into the water. In other words, Maria considers the turn dangerous if she immediately gets into the water if it is ignored.

Help Maria get ready for the competition — determine the number of dangerous turns on the track.

Input
The first line of the input contains an integer n (4 ≤ n ≤ 1000) — the number of straight sections of the track.

The following (n + 1)-th line contains pairs of integers (xi, yi) ( - 10 000 ≤ xi, yi ≤ 10 000). The first of these points is the starting position. The i-th straight section of the track begins at the point (xi, yi) and ends at the point (xi + 1, yi + 1).

It is guaranteed that:

the first straight section is directed to the north;
the southernmost (and if there are several, then the most western of among them) point of the track is the first point;
the last point coincides with the first one (i.e., the start position);
any pair of straight sections of the track has no shared points (except for the neighboring ones, they share exactly one point);
no pair of points (except for the first and last one) is the same;
no two adjacent straight sections are directed in the same direction or in opposite directions.
Output
Print a single integer — the number of dangerous turns on the track.

input
6
0 0
0 1
1 1
1 2
2 2
2 0
0 0
output
1
input
16
1 1
1 5
3 5
3 7
2 7
2 9
6 9
6 7
5 7
5 3
4 3
4 4
3 4
3 2
5 2
5 1
1 1
output
6

题目链接：cf-659D

题目思路：可知，凹点会掉下去。假设初始为一个矩形，每次增加一个凹或者凸，一定会增加4个点，并且两个凸点两个凹点。所以答案为(n - 4)/2

以下是代码：

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
int main()
{
    int n;
    cin >> n;
    cout << (n - 4) / 2 << endl;
    return 0;
}